Given a real vector space $V$ with an almost complex structure $I$, and an inner product $\langle, \rangle$. Can define a fundamental form $(,):= \langle I(),() \rangle$. The claim is that is real of type $(1,1)$, i.e. $(,) \in \wedge^2 V^{\ast} \cap \wedge^{(1,1)}V^{\ast}$.
Why should it be real? It does not make sense at all, since if I take for instance $iv \in V_{\mathbb C}$, $v, w \in V$, then $(w, iv) = i(I(w), v)$ which is complex.
Also, why is it denoted by $\wedge^2 V^{\ast} \cap \wedge^{(1,1)}V^{\ast}$? Isn't $\wedge^{(1,1)}V^{\ast}$ a subspace of $\wedge^{2}V^{\ast}$?
This statement is taken from Introduction to Complex Geometry Lemma1.2.24, can someone clarify what it means by saying that the form is real?
This matter has also puzzled me for a long time.
$ V^*_{\mathbb{C}} $, the complexification of the dual space of $ V $, has a direct sum decomposition $$V^*_\mathbb{C}=W^{1,0}\oplus W^{0,1}$$ since on $ V^* $ there is a natural complex structure induced by $ I $. Then we have $$ \bigwedge^2 V^*_\mathbb{C}=\bigwedge^2 W^{1,0}\oplus W^{1,1}\oplus \bigwedge^2W^{0,1}, $$ where $ W^{1,1} $ is the image of $$ \varphi\colon W^{1,0}\otimes W^{0,1}\longrightarrow \bigwedge^2 V^*_\mathbb{C},\quad f\otimes g\longmapsto f\wedge g. $$
Besides, for every $ \omega\in \bigwedge^2 V^* $, which can be regarded as an alternating function $$ \omega\colon V\times V\longrightarrow \mathbb{R}, $$ can be extended by linear to (still denoted by $ \omega $) $$ \omega\colon V_\mathbb{C}\times V_\mathbb{C}\longrightarrow \mathbb{C}. $$ In this way we embed $ \bigwedge^2 V^* $ into $ \bigwedge^2 V^*_{\mathbb{C}} $. Now we can talk about the intersection $$ W^{1,1}_\mathbb{R}:=W^{1,1}\cap \bigwedge^2V^*\subset \bigwedge^2 V^*_{\mathbb{C}}. $$ We say elements of $ W^{1,1}_\mathbb{R} $ are alternating real form of type $ (1,1) $. Now it is clear how the name "real" and " type $ (1,1) $ " come from.
My references
Warnning: The notation in this answer such as $ W^{1,1}_\mathbb{R} $ may not be standard.